The basis of a vector space -- Abstract Linear Algebra 10

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[malawigw]

I love this series so much! Professor Benedict Gross once said "you can never study too much linear algebra" and I believe he is correct on that!

[BethKjos]

Proposition around @20:00 seems like just a restatement of the definition of linearly dependence: There's a vector of coefficients we call "alpha"... Scale it by the negative multiplicative inverse of a non-zero component then add that component's vector to both sides.

[StanleyDevastating]

Warm up problems are fun. For the first one I got that B was a basis for F_3 and not a basis for F_5. Also 4 = 1 and 5=-1 in F_3, so (4 5) is just a basis vector, so I think gamma_B (4 5) = (0 1) in F_3.

[nathanisbored]

its worth noting that, with only two vectors you can tell if they are independent or not by checking if one is a multiple of the other. in fact, linear dependence is kind of just a generalization of "something is a multiple of another thing" but for when you have more than two things. thats probably why you didnt give any examples like that, because they arent as interesting, but for a topic so abstract, i think an example where you can just eyeball the answer intuitively is still instructive.

[wafelsen]

10:30 the best way I saw was to take the middle equation and subtract the other two from it. That neatly gets just -y=0 and it is clear from there.

[yakovify]

7:20 you say there are infinitely many solutions but I believe that only applies to infinite fields and since this field is finite there's is a finite amount. In fact since the space of solutions is one dimensional there are exactly 5 solutions.

[oliverdixon]

For a slightly simpler presentation of the first proposition's proof, could you say that since a reordered linearly dependent list is still linearly dependent, WLOG suppose that $ \alpha_n \neq 0 $, and then state that $ v_n = -1/\alpha_n ( \alpha_1 v_1 + \ldots + \alpha_{n-1} v_{n-1} ) $? (I'm just trying to check my own understanding here.)

[ingobojak]

The calculation at 16:14 is not necessary. Since the equation has to be true for all x, the coefficients of all powers of x have to be zero separately. And in this case the coefficients of the first and second power of x immediately deliver the result that the alphas have to be zero.

[iooooooo]

23:24 For this argument to work I think one technically needs to prove that -1 != 0 for any field k. Which is pretty easy, but I'm not sure it was explicitly called out in this course yet. (Though I think we might have seen 1 != 0, which gets us there pretty easily.)

[artificialresearching]

P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)